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Re: time serie -> sum inside a loop

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46138] Re: time serie -> sum inside a loop
  • From: bobhanlon at aol.com (Bob Hanlon)
  • Date: Sat, 7 Feb 2004 04:02:47 -0500 (EST)
  • References: <bvvneh$j53$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

n=4;
a = ToExpression[Table["a"<>ToString[k],{k,n}]]

{a1, a2, a3, a4}

mu=Tr[a]/Length[a]

(1/4)*(a1 + a2 + a3 + a4)

Y = Rest[FoldList[Plus, 0, a-mu]]

{a1 + (1/4)*(-a1 - a2 - a3 - a4),
  a1 + a2 + (1/2)*(-a1 - a2 - a3 - a4),
  a1 + a2 + a3 + (3/4)*(-a1 - a2 - a3 - a4), 0}

i=Random[Integer, {1, n}];
Y[[i]] == Sum[a[[k]]-mu, {k,i}]

True


Bob Hanlon

In article <bvvneh$j53$1 at smc.vnet.net>, "paolo" <tarpanelli at libero.it> wrote:

<< Given a time serie
a:= {a[[1]],a[[2]],a[[3]],...,a[[n]]}

how can i solve this loop with mathematica?

-------------
Y[[i]]:=Sum_(for k=1 to i) [a[[k]]-mu]
-------------

where mu is the mean of time serie a.


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